[leetcode]95. Unique Binary Search Trees II

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,Given n = 3, your program should return all 5 unique BST's shown below.

 1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

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解题思路：
本题是96.Unique Binary Search Trees的延伸，看网上大神么的解题思路，写的都是“有了96题的解题思路，本题应该不难”，然后给出代码。作为编程小菜的我，想了半天，还是没想出来（心中飘过千万只CNM)。

看完答案以后，总结一下，两道题的核心思想确实一样，都是分治法，但是解题的着眼点真的不一样，96.Unique Binary Search Trees的核心是用DP来求出树的个数，而本题的着眼点是求出所有的树，关键在于如何用递归的方法生成所有的树，如果对递归不熟悉，对如何生成树不熟悉，真的是无从下手。

具体代码如下：

class Solution {
public:
    vector genBST(int begin, int end)
    {
        vector ret;
        if(begin > end) return ret;
        for(int i = begin; i <= end; ++i)
        {
            vector l = genBST(begin, i-1);
            vector r = genBST(i+1,end);
            
            int lSz = l.empty() ? 1 : l.size();
            int rSz = r.empty() ? 1 : r.size();
            
            for(int j = 0; j < lSz; ++j)
            {
                for(int k = 0; k < rSz; ++k)
                {
                    TreeNode* root = new TreeNode(i);
                    if(l.empty()) root->left =NULL;
                    else root->left = l[j];
                    
                    if(r.empty()) root->right = NULL;
                    else root->right = r[k];
                    
                    root-> val = i;
                    ret.push_back(root);
                }
            }
        }
        return ret;
     }
    vector generateTrees(int n) {
        return genBST(1,n);
    }
};